
Together with Theorem A.4 of [1]}, this leads to the following decomposition of \(s_n(z)\) :
\(\phantom{\,.} s_n(z) &= \frac{1}{n}\sum _{k=1}^n \tilde{b}_k = \bar{b}_n + \frac{1}{n} \sum _{k=1}^n ( -\bar{b}_n^2 \bar{\varepsilon }_k + \bar{b}_n^2 \tilde{b}_k \bar{\varepsilon }_k^2 )  \,.\) 
